Roboguru

Buktikan bahwa a) (f∘g)−1(x)=(g−1∘f−1)(x) untuk f(x)=3x+7g(x)=9−x

Pertanyaan

Buktikan bahwa

a) begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end style

untuk

begin mathsize 14px style f open parentheses x close parentheses equals 3 x plus 7 g open parentheses x close parentheses equals 9 minus x end style 

Pembahasan Video:

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell 3 open parentheses g open parentheses x close parentheses close parentheses plus 7 end cell row blank equals cell 3 open parentheses 9 minus x close parentheses plus 7 end cell row blank equals cell 27 minus 3 x plus 7 end cell row blank equals cell 34 minus 3 x end cell end table end style 

Misalkan begin mathsize 14px style y equals open parentheses f ring operator g close parentheses open parentheses x close parentheses end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 34 minus 3 x end cell row cell 3 x end cell equals cell 34 minus y end cell row x equals cell fraction numerator 34 minus y over denominator 3 end fraction end cell end table end style

Jadi, diperoleh invers dari fungsi komposisi yaitu

begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 34 minus x over denominator 3 end fraction end style

Kemudian dicari invers dari masing masing fungsi, yaitu

  • Misalkan begin mathsize 14px style f open parentheses x close parentheses equals y end style dan begin mathsize 14px style g open parentheses x close parentheses equals y end style 

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 Diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell equals cell 9 minus f to the power of negative 1 end exponent open parentheses x close parentheses end cell row blank equals cell 9 minus fraction numerator x minus 7 over denominator 3 end fraction end cell row blank equals cell fraction numerator 27 minus x plus 7 over denominator 3 end fraction end cell row cell left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis end cell equals cell fraction numerator 34 minus x over denominator 3 end fraction end cell end table end style

 Jadi, terbukti begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end style

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Terakhir diupdate 06 Oktober 2021

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Jika f(x)=4x−1 dan g(x)=2x−3, nilai (f−1∘g)−1(2)adalah...

Pembahasan Soal:

Ingat kembali sifat invers fungsi komposisi berikut.

(f1)1(x)(fg)1(x)==f(x)(f1g1)(x)

Dengan menggunakan konsep invers fungsi. Misalkan begin mathsize 14px style g open parentheses x close parentheses equals y end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g open parentheses x close parentheses end cell equals y row cell 2 x minus 3 end cell equals y row cell 2 x end cell equals cell y plus 3 end cell row x equals cell fraction numerator y plus 3 over denominator 2 end fraction end cell row cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator x plus 3 over denominator 2 end fraction end cell end table end style

Selanjutnya dengan menggunakan sifat invers dari kompisisi fungsi diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f to the power of negative 1 end exponent ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell g to the power of negative 1 end exponent open parentheses x close parentheses ring operator open parentheses f to the power of negative 1 end exponent close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses x close parentheses ring operator f open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses f open parentheses x close parentheses close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses 4 x minus 1 close parentheses end cell row blank equals cell fraction numerator 4 x minus 1 plus 3 over denominator 2 end fraction end cell row blank equals cell fraction numerator 4 x plus 2 over denominator 2 end fraction end cell row cell open parentheses f to the power of negative 1 end exponent ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end cell equals cell fraction numerator 4 open parentheses 2 close parentheses plus 2 over denominator 2 end fraction end cell row blank equals cell 10 over 2 end cell row blank equals 5 end table end style

Dengan demikian nilai dari begin mathsize 14px style open parentheses f to the power of negative 1 end exponent ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end style adalah begin mathsize 14px style 5 end style.

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Diketahui f(x)=1+3x−8x−1​​ dengan x=8. Jika f−1(b)=9, maka nilai b yang memenuhi adalah …

Pembahasan Soal:

begin mathsize 14px style f open parentheses x close parentheses equals 1 plus cube root of fraction numerator x minus 1 over denominator x minus 8 end fraction end root end style dengan begin mathsize 14px style x not equal to 8 end style. Jika begin mathsize 14px style f to the power of negative 1 end exponent open parentheses b close parentheses equals 9 end style, maka begin mathsize 14px style f to the power of negative 1 end exponent left parenthesis b right parenthesis equals 9 space maka space f open parentheses 9 close parentheses equals b end style. Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row b equals cell f open parentheses 9 close parentheses end cell row b equals cell 1 plus cube root of fraction numerator 9 minus 1 over denominator 9 minus 8 end fraction end root end cell row b equals cell 1 plus cube root of 8 end cell row b equals cell 1 plus 2 end cell row b equals 3 end table end style 

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Diketahui f(x) = 4x + 2 dan g(x)=x+1x−3​,x=−1. Invers dari (g o f)(x) adalah ....

Pembahasan Soal:

begin mathsize 14px style open parentheses straight g ring operator straight f close parentheses open parentheses straight x close parentheses equals straight g open parentheses straight f open parentheses straight x close parentheses close parentheses  equals straight g open parentheses 4 straight x plus 2 close parentheses  equals fraction numerator left parenthesis 4 straight x plus 2 right parenthesis minus 3 over denominator left parenthesis 4 straight x plus 2 right parenthesis plus 1 end fraction  equals fraction numerator 4 straight x minus 1 over denominator 4 straight x plus 3 end fraction    Invers space dari space straight h left parenthesis straight x right parenthesis equals fraction numerator ax plus straight b over denominator cx plus straight d end fraction space adalah space straight h to the power of negative 1 end exponent left parenthesis straight x right parenthesis equals fraction numerator negative dx plus straight b over denominator cx minus straight a end fraction  Sehingga space invers space dari space left parenthesis straight g ring operator straight f right parenthesis left parenthesis straight x right parenthesis equals fraction numerator 4 straight x minus 1 over denominator 4 straight x plus 3 end fraction space adalah  open parentheses straight g ring operator straight f close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator negative 3 straight x minus 1 over denominator 4 straight x minus 4 end fraction  open parentheses straight g ring operator straight f close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator 3 straight x plus 1 over denominator 4 minus 4 straight x end fraction end style

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Diketahui f(x)=7x−35−4x​. Bila f−1(x) adalah invers dari f(x), maka f−1(x)=...

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell fraction numerator a x plus b over denominator c x plus d end fraction rightwards double arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator negative d x plus b over denominator c x minus a end fraction end cell end table end style
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell f left parenthesis x right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 5 minus 4 x over denominator 7 x minus 3 end fraction end cell rightwards double arrow cell f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 3 x plus 5 over denominator 7 x plus 4 end fraction end cell end table end style

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Jika /(x1-1)--3, maka nilai ri (4) , maka nilai 1(-2 x-1) x+3 adalah

Pembahasan Soal:

Jika f to the power of left parenthesis negative 1 right parenthesis end exponent left parenthesis a right parenthesis equals b maka f(b)=a

Misal f to the power of left parenthesis negative 1 right parenthesis end exponent left parenthesis negative 2 right parenthesis equals p, maka :

f(p) = -2

Sedangkan
f open parentheses fraction numerator 1 over denominator x minus 1 end fraction close parentheses equals fraction numerator x minus 6 over denominator x plus 3 end fraction

Jadi, dapat disimpulkan:
1) negative 2 equals fraction numerator x minus 6 over denominator x plus 3 end fraction  minus 2 x minus 6 equals x minus 6 space  minus 3 x equals 0 space  x equals 0

2) p space equals space fraction numerator 1 over denominator x minus 1 end fraction  p equals fraction numerator 1 over denominator 0 minus 2 end fraction  equals negative 1

 

 

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